Question

If P is the point in the argand diagram corresponding to the complex number $\sqrt{3}+i$ and if $OPQ$ is an isosceles right angled triangle ,right angled at $O$, then $Q$ represents the complex number:

Answer 1

Hint: The argand diagram is used for graphical representation of complex numbers in the form of $x+iy$ in the complex plane. Similar to $x-axis$ and $y-axis$ in the two dimensional geometry we have a horizontal axis used to indicate real numbers and a vertical axis used to represent imaginary numbers in case of an argand plane. For example: $5+4i$ represents the ordered pair $\left(5,4\right)$ geographically in the argand plane.

Complete step-by-step solution:
The given complex number is; $P=\sqrt{3}+i$
According to the given question, let us draw the diagram to understand the question in a better way;

Figure $\left(1\right)$ : Isosceles right angled triangle $OPQ$
$\Rightarrow z=x+iy\left(\text{Real part = }x,\text{ Imaginary part = }y\right)$
We know that if two lines are perpendicular to each other then the product of their slopes will be equal to $-1$ ; i.e.
$\Rightarrow m_1\times m_2=-1$
First, let us calculate the slope $\left(m_1\right)$ of line OP;
$\Rightarrow m_1=\left[{\displaystyle \frac{1-0}{\sqrt{3}-0}}\right]={\displaystyle \frac{1}{\sqrt{3}}}......\left(1\right)$
Now, let us calculate slope $\left(m_2\right)$ of line OQ;
$\Rightarrow m_2=\left[{\displaystyle \frac{y-0}{x-0}}\right]={\displaystyle \frac{y}{x}}......\left(2\right)$
Put the respective values in the property $m_1\times m_2=-1$ , we get;
$\Rightarrow {\displaystyle \frac{1}{\sqrt{3}}}\times {\displaystyle \frac{y}{x}}=-1$
  Simplifying the above equation;
$\Rightarrow y=-\sqrt{3}x......\left(3\right)$
Using the properties of isosceles triangles (stated in the note part) let us try to solve our question;
According to figure $\left(1\right)$ , $OP=OQ\left(\because \text{Congruent sides of the isosceles triangle}\right)$
$\therefore O{P}^2=O{Q}^2$ will also be true.
By the distance formula between two points, we know that;
$\Rightarrow d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ ( where $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are the coordinates of first
point and second point respectively )
$\Rightarrow O{P}^2=\left[\left({\left(\sqrt{3}\right)}^2-0\right)+\left({\left(1\right)}^2-0\right)\right]$
$\Rightarrow O{P}^2=3+1=4......\left(4\right)$
Now, let us similarly find the equation for OQ;
$\Rightarrow O{Q}^2=\left[\left({\left(x\right)}^2-0\right)+\left({\left(y\right)}^2-0\right)\right]$
$\Rightarrow O{Q}^2=x^2+y^2$
Now, put the value of $y=-\sqrt{3}x$ from equation $\left(3\right)$ in the above equation we get;
$\Rightarrow O{Q}^2=x^2+{\left(-\sqrt{3}x\right)}^2$
The above equation can be further simplified as;
$\Rightarrow O{Q}^2=x^2+3x^2$
$\Rightarrow O{Q}^2=4x^2......\left(5\right)$
On comparing equation $\left(4\right)$ and equation $\left(5\right)$ , we get;
$\Rightarrow 4x^2=4$
$\Rightarrow x^2=1$
Which means ; $x=\pm 1$
Now put the value of $x\text{ in equation }\left(3\right)$ to get the value of $y$ ; we get two cases;
When $x=1\text{ , }y=-\sqrt{3}$
And $x=-1y=\sqrt{3}$
Therefore, there can be two possible values of $Q$ , i.e. $Q\left(\pm 1,\mp \sqrt{3}\right)$
Therefore, the answer for this question is $$Q=1-\sqrt{3}i\text{ or }Q=-1+\sqrt{3}i$$.

Note: Here are the important properties of a right angled isosceles triangle. It will be easier to understand via a diagram, the isosceles triangle theorem states that;

Figure $\left(2\right)$ : Isosceles triangle theorem
$\left(1\right)$ In the above diagram $AB=BC=X$ , means two sides of the triangle are congruent, then the third side will be equal to $X\sqrt{2}$ means the hypotenuse is $\sqrt{2}$ times the length of a leg.
$\left(2\right)$ If $AB=BC$ then $\mathrm{\angle }BAC=\mathrm{\angle }BCA$ .